for \(v\in S\) , if \(d(v) > 2\) , then there is at most one negative edge incident at \(v\)  in
\(S\) .
Theorem 2.8 
Let \(\Sigma =(\Gamma , \sigma )\) , where
\(\Gamma ={\mathrm {Cay}}(\mathbb {Z}_{p_{1}}\times \mathbb {Z}_{p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}} \ldots p_{k}^{\alpha _{k}}}, \Phi )\) .
Then \(L(\Sigma )\)  is balanced if and only if one of the prime factors is 2.
Necessity: Let \(L(\Sigma )\)  is balanced. Assume that the conclusion is false. Suppose \(p_{i}\ge 3\) 
for any \( i=1, 2, \ldots , k\) . It is easy to see that vertex \((0, 1)\)  is adjacent to vertices
\((1, 0), (2, 0), \ldots , (p_{1}-1, 0)\) . By definition \(\Phi \) , the vertices \((0, 1), (1, 0), (2, 0), \ldots , (p_{1}-1, 0)\)  are not in
\(\Phi \) . Hence \(d^{-}((0, 1))\ge 2\) . By the Theorem REF , condition (ii) does not hold. Therefore
\(L(\Sigma )\)  is unbalansed. Which is in contradiction with the assumption. Thus one of the prime factors is 2.
Sufficiency: Suppose one of the prime factors is 2. By Theorem REF , \(\Sigma \)  is
balanced so, according to Lemma 1.5[1]} and Theorem 1.6[1]}, \(\Sigma \)  is switching
equivalent to a graph whose all edges are positive. Hence \(\Sigma \)  can be considered all-positive.
Therefore \(L(\Sigma )\)  is balanced.

Clusterability and sign-compatibility of \(\Sigma \)
In this chapter two properties namely clusterability and sign-compatibility for \(\Sigma \)  are investigated.
Theorem 3.1 [3]}
Let \(S\)  be any signed graph. Then \(S\)  has a clustering if and only if \(S\)  contains
no cycle having exactly one negative line.

Theorem 3.2 [4]}
A sigraph \(S\)  is sign-compatible if and only if \(S\)  does not contain a subsigraph
that is isomorphic to either of the following two sigraphs formed from the path graph
\(P4 = (x, u, v, y)\) ; \(S1\)  with both edges \(xu\)  and \(vy\)  negative and the edge \(uv\)  positive and \(S2\) 
obtained from \(S1\)  by identifying the vertices \(x\)  and \(y\)  (see Figure REF  ).

<FIGURE>Theorem 3.3 
Let \(\Sigma =(\Gamma , \sigma )\) , where
\(\Gamma ={\mathrm {Cay}}(\mathbb {Z}_{p_{1}}\times \mathbb {Z}_{p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}} \ldots p_{k}^{\alpha _{k}}}, \Phi )\) .
Then \(\Sigma \)  is clusterable if and only if \(\Sigma \)  is balanced.
Nesssity: Suppose \(\Sigma \)  is clusterable. Assume that the conclusion is false.
Suppose \(\Sigma \)  is unbalanced. Using Theorem REF , \(p_{i}\ge 3\)  for any
\( i=1, 2, \ldots , k\) . Because \(p_{i}\ge 3\)  so \((2, 2)\in \Phi \)  and also it is adjacent to
vertices \((0, 1)\)  and \((1, 0)\) . We now consider the cycle \(C=((0, 1), (2, 2), (1, 0), (0, 1))\) 
in \(\Sigma \) . By the definition of \(\sigma \) , we have \(\sigma ((0, 1)(2, 2))=\sigma ((2, 2)(1, 0))=+\) 
and \(\sigma ((1, 0)(0, 1))=-\) . Hence \(C\)  is a cycle with exactly one negative edge. Therefore, according
to Theorem REF , \(\Sigma \)  can not be clusterable, a contradiction to the hypothesis.
Hence one of the prime factors is 2 so \(\Sigma \)  is balanced.
Sufficiency: Suppose \(\Sigma \)  is balanced. Thus all cycles of \(\Sigma \)  are positive.
So the number of negative edges in them is even. Hence it can not include the cycle with a single
negative edge. Therefore, according to Theorem REF , \(\Sigma \)  is clusterable.

Theorem 3.4 Let \(\Sigma =(\Gamma , \sigma )\) , where
\(\Gamma ={\mathrm {Cay}}(\mathbb {Z}_{p_{1}}\times \mathbb {Z}_{p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}} \ldots p_{k}^{\alpha _{k}}}, \Phi )\)  and \(p_{i}\ge 2\)  for any
\( i=1, 2, \ldots , k\) . Then \(\Sigma \)  is sign-compatible.
Assume \(\Sigma \)  is sign-incompatible. Then by Theorem REF , \(\Sigma \)  contains a subsigraph
isomorphic to either \(S1\)  or \(S2\) . Let \(P=((u, v), (u^{^{\prime }}, v^{^{\prime }}), (u^{^{\prime \prime }}, v^{^{\prime \prime }}), (u^{^{\prime \prime \prime }}, v^{^{\prime \prime \prime }}))\)  is
a path in \(\Sigma \)  such that \(\sigma ((u, v)(u^{^{\prime }}, v^{^{\prime }}))=\sigma ((u^{^{\prime \prime }}, v^{^{\prime \prime }})(u^{^{\prime \prime \prime }}, v^{^{\prime \prime \prime }}))=-\)  and
\(\sigma ((u^{^{\prime }}, v^{^{\prime }})(u^{^{\prime \prime }}, v^{^{\prime \prime }}))=+\) . According to the definition of \(\sigma \) , since
\((u^{^{\prime }}, v^{^{\prime }})(u^{^{\prime \prime }}, v^{^{\prime \prime }})\)  is positive edge hence two cases may occure.
If \((u^{^{\prime }}, v^{^{\prime }})\)  and \((u^{^{\prime \prime }}, v^{^{\prime \prime }})\)  belong to \(\Phi \)  then all the edges on \(P\)  are positive,
a contradiction to hypothesis. Now if one of the vertices \((u^{^{\prime }}, v^{^{\prime }})\)  and \((u^{^{\prime \prime }}, v^{^{\prime \prime }})\) 
belong to \(\Phi \) . Without loss of generality assume that \((u^{^{\prime }}, v^{^{\prime }})\in \Phi \)  and \((u^{^{\prime \prime }}, v^{^{\prime \prime }})\notin \Phi \) .
Then we have \(\sigma ((u, v)(u^{^{\prime }}, v^{^{\prime }}))=((u^{^{\prime }}, v^{^{\prime }})(u^{^{\prime \prime }}, v^{^{\prime \prime }}))=+\)  and
\(\sigma ((u^{^{\prime \prime }}, v^{^{\prime \prime }})(u^{^{\prime \prime \prime }}, v^{^{\prime \prime \prime }}))=-\) . Which is contradiction the hyposesis. Thus
\(\Sigma \)  does not contain a subsigraph isomorfic to \(S1\) .
Now let \(\Sigma \)  contains a subsigraph isomorphic to \(S2\) . As proved for \(S1\) ,
according to the definition of \(\sigma \) , no positive edge can be placed between two
negative edges. Hence \(\Sigma \)  does not contain a subsigraph isomorfic to \(S2\) .
Therefore \(\Sigma \)  is sign-compatible.

